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What is the smallest number which when divided by 72 and 96, respectively, leaves a remainder 5 in each case?
283
571
581
293
- To solve the problem, understand it's about finding a least common number (LCM issue) with a remainder pattern.
- You need the smallest number \( x \) such that:
- \( x \equiv 5 \pmod{72} \)
- \( x \equiv 5 \pmod{96} \)
- In simpler terms: When you subtract 5 from this number, it should be divisible by both 72 and 96.
- LCM of 72 and 96 helps here, which is 288.
- Add 5 to a multiple of this LCM for correct remainder.
- Options checked:
- Option 1: 283 — Subtracting 5 gives 278, not a solution.
- Option 2: 571 — Subtracting 5 gives 566, not a solution.
- Option 3: 581 — Subtracting 5 gives 576, a multiple of 288.
- Option 4: 293 — Subtracting 5 gives 288, a multiple of 288.
- Correct answer is Option 4: 293.
.
By: santosh ProfileResourcesReport error
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